Question: A particle moves in the $xy$ -plane so that at any time $t\geq 0$ its position vector is $(2t^2+5,-t^3+4t^2)$. What is the magnitude of the particle's acceleration vector at $t=1$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\sqrt{29}$ (Choice B) B $2\sqrt{5}$ (Choice C) C $\sqrt{58}$ (Choice D) D $4$
Solution: Background When solving motion problems, it's important to remember the relationship between the position vector $(x,y)$, the velocity vector $\vec{v}(t)$, and the acceleration vector $\vec{a}(t)$ of the moving particle: $\vec{v}(t)=\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)$ $\vec{a}(t)=\dfrac{d}{dt}\vec{v}(t)=\left(\dfrac{d^2x}{dt^2},\dfrac{d^2y}{dt^2}\right)$ Setting up the math We are given that the particle's position vector is $(2t^2+5,-t^3+4t^2)$. We are asked to find the magnitude of the particle's acceleration vector at $t=3$. In other words, we need to find $||\vec{a}(3)||$. Finding $\vec{v}(t)$ $\begin{aligned} \vec{v}(t)&=\left(\dfrac{d}{dt}(2t^2+5),\dfrac{d}{dt}(-t^3+4t^2)\right) \\\\ &=(4t,-3t^2+8t) \end{aligned}$ Finding $\vec{a}(t)$ $\begin{aligned} \vec{a}(t)&=\dfrac{d}{dt}\vec{v}(t) \\\\ &=\left(\dfrac{d}{dt}(4t),\dfrac{d}{dt}(-3t^2+8t)\right) \\\\ &=(4,-6t+8) \end{aligned}$ Finding $\vec{a}(1)$ $\begin{aligned} \vec{a}({1})&=(4,-6({1})+8) \\\\ &=(4,2) \end{aligned}$ Finding $||\vec{a}(1)||$ $\begin{aligned} ||\vec{a}(1)||&=||(C{4},{2})|| \\\\ &=\sqrt{(C{4})^2+({2})^2} \\\\ &=\sqrt{20} \\\\ &=2\sqrt5 \end{aligned}$ In conclusion, the magnitude of the particle's acceleration vector at $t=1$ is $2\sqrt5$.